Determine the equation of the inverse function if f(x) = 2x^2+3, and x≥0.
The answer is supposed to be f^-1(x)=[√(2x-6)]/2.
This is what I did:
x=2y^2+3
2y^2=x-3
y^2=(x-3)/2
y=√[(x-3)/2]
Did I do something wrong?
Thanks!
I think you did it correctly.
Of course the domain is limited because the answer is imaginary if x <3
the answers are the same
√[(x-3)/2] = √[(x-3)/2 * 2/2] = √(2x-6)/2
To find the inverse of a function, you typically want to solve for the input variable in terms of the output variable. In this case, we're trying to find the inverse of the function f(x) = 2x^2 + 3.
Your working is mostly correct, but let's go through the steps together to make sure we didn't miss anything.
Step 1: Replace f(x) with y to rewrite the equation as: y = 2x^2 + 3.
Step 2: Swap the variables x and y: x = 2y^2 + 3.
Step 3: Solve for y.
- Subtract 3 from both sides: x - 3 = 2y^2.
- Divide both sides by 2: (x - 3)/2 = y^2.
- Take the square root of both sides: √((x - 3)/2) = y.
Now, here's where the discrepancy arises. To ensure that the inverse function is valid, we need to verify the domain and range restrictions.
Given that the original function has the condition x ≥ 0, we must impose similar restrictions on the inverse function. The square root term (√((x - 3)/2)) is only defined for (x - 3)/2 ≥ 0.
Thus, we can isolate this condition as (x - 3) ≥ 0, which gives x ≥ 3.
Therefore, the inverse function is:
f^(-1)(x) = √((x - 3)/2)
= [√(x - 3)] / √2.
However, note that, in the final step, we rationalize the denominator by multiplying both the numerator and denominator by √2:
f^(-1)(x) = [√2 √(x - 3)] / [√2 (√2)]
= [√(2x - 6)] / 2.
So, the correct equation for the inverse function is indeed f^(-1)(x) = [√(2x - 6)] / 2.
You made a minor mistake by not rationalizing the denominator, but apart from that, your steps were correct. Just remember to check any domain and range restrictions when finding the inverse function.
I hope this explanation clears up any confusion and helps you understand the solution! Let me know if you have any other questions.